Circle Inscribed - Triangle Problems

Category: Plane Geometry

"Published in Newark, California, USA"

The base of an isosceles triangle is 16 in. and the altitude is 15 in. Find the radius of the inscribed circle.

Solution:

To illustrate the problem, it is better to draw the figure as follows

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The area of ∆ABC is

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By using Pythagorean Theorem, we can solve for the two legs of an isosceles triangle as follows

Next, draw the angle bisectors of an isosceles traingle as follows

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The intersection of the angle bisectors of an isosceles triangle is the center of an inscribed circle which is point O. From point O, draw a line which is perpendicular to AB, draw a line which is perpendicular to AC, and draw a line which is perpendicular to BC. These three lines will be the radius of a circle. 

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The total area of an isosceles triangle is equal to the area of three triangles whose vertex is point O. Therefore, the radius of an inscribed circle is

Regular Polygon Problems

Category: Plane Geometry

"Published in Vacaville, California, USA"

Find the perimeter of a regular hexagon that is inscribed in a circle of radius 8 m.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

If you will draw the diagonals that passes through the center of a circle, there are six triangles in a regular hexagon as follows

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The center of a circle bisects the three diagonals of a regular hexagon. The bisected diagonals are equal to the radius of a circle which is 8 m. The vertex angle of each triangles can be calculated as follows

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The triangles in hexagon are isosceles triangles because the two sides of each triangles are congruent which is 8 m. If the two sides of an isosceles triangle are congruent, then its base angles are congruent also. We can calculate the base angle of an isosceles triangle as follows

Since the base angles of an isosceles triangle are the same as the vertex angle which is 60°, then the isosceles triangle is an equiangular or equilateral triangle. Hence, the sides of an hexagon is x = 8 m. 

Therefore, the perimeter of a regular hexagon is

Proving - Inscribed Triangle, Circle

Category: Plane Geometry

"Published in Newark, California, USA"

Given: ∆ABC inscribed in circle O

Prove: m∠BAC = ½ m(arc BC)

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Solution:

Consider the given figure and we will label further as follows

Photo by Math Principles in Everyday Life

Proof:

1. Statement: From point A, draw a line that passes through the center of a circle.

    Reason: Line AO will be used later to get m∠BOC.

2. Statement: Draw radius OB and let m∠BAO = x.

    Reason: Radius OB will be used to create an isosceles ∆BOA and x as the base angle.

3. Statement: OB ≅ OA and m∠BAO ≅ m∠OBA.

    Reason: OB and OA are the radius of a circle and the two sides of an isosceles triangles are congruent. The two base angles of an isosceles triangles are congruent.

4. Statement: m∠BOA = 180° - 2x

    Reason: The sum of the three angles of a triangle is always equal to 180°.

5. Statement: m∠BOD = 180° - (180° - 2x) = 2x

    Reason: The sum of supplementary angles is always 180°. m∠BOA and m∠BOD are supplementary angles.

6. Statement: Draw radius OC and let m∠CAO = y.

    Reason: Radius OC will be used to create an isosceles ∆COA and y as the base angle.

7. Statement: OC ≅ OA and m∠CAO ≅ m∠OCA.

    Reason: OC and OA are the radius of a circle and the two sides of an isosceles triangles are congruent. The two base angles of an isosceles triangles are congruent.

8. Statement: m∠COA = 180° - 2y

    Reason: The sum of the three angles of a triangle is always equal to 180°.

9. Statement: m∠COD = 180° - (180° - 2y) = 2y

    Reason: The sum of supplementary angles is always 180°. m∠COA and m∠COD are supplementary angles.

10. Statement: m∠BAC = x + y and m∠BOC = 2x + 2y.

      Reason: Addition property of angles

11. Statement: m∠BAC = ½ m∠BOC

      Reason: If OA, OB, and OC are equidistant from its center at O, then m∠BAC = ½ m∠BOC.

12. Statement: m∠BOC = m(arc BC)

      Reasons: If m∠BOC is a central angle formed by the two radii OB and OC, then m(arc BC) is an arc formed by a central angle. 

13. Statement: m∠BAC = ½ m∠BOC = ½ m(arc BC)

      Reasons: Substitution proposition.